# Classify all Maximal Ideals on this Ring

• Oster
In summary: J(y). Therefore, there cannot exist any other maximal ideals in C[0,1] besides the J(y)'s.In summary, the maximal ideals in the commutative ring of continuous functions on [0,1] (C[0,1],+,.) are the sets J(y) = {f in C[0,1] / f(y) = 0} for all y in [0,1]. These are the only maximal ideals and any other ideal in C[0,1] is either the zero ideal or a subset of one of the J(y)'s.
Oster
Classify all maximal ideals on the commutative ring of continuous functions on [0,1] (C[0,1],+,.)

I am not too confident about my solution. Can someone please look through it?

Fortunate guess:-
The set J(y) = {f in C[0,1] / f(y)=0} where y is fixed in [0,1]

This is an ideal because
(1) If g,h belong to J(y) then g-h belongs to J(y) => (J(y),+) is a subgroup of (C[0,1],+)
(2) If g is in J(y) and h is in C[0,1] then g.h is in J(y)

Also, C[0,1]/J(y) is a field because,
If J(y)+f is in C[0,1]/J(y) and J(y)+f =/= J(y), then f(y) =/= 0.
Let g(x)=1/f(y) for all x in C[0,1]
Then, (J(y)+g).(J(y)+f)=(J(y)+f.g)=(J(y)+1) [because f.g-1 is in J(y)]
Hence (J(y)+f) is invertible.

And, C[0,1]/J(y) is a field => J(y) is a maximal ideal.

To show there are no other maximal ideals:-

Let I be an ideal of C[0,1] such that I is not {0} and I is not a subset of any J(y)

That implies, for each z in [0,1], there exists f in I such that f(z) =/= 0.
multiplying this f by an appropriate constant function AND an appropriate continuous piecewise linear function will guarantee, for each z, the existence of a function g in I such that g(z)>1 AND g>0 on [0,1].

for z in [0,1], the corresponding function g and e=1/2, there exists a number d(z)>0 such that in B(z,d(z)) g is always greater than 1/2.

B(z,d(z)) is an open cover for [0,1] so it must admit a finite subcover. This will give me a finite number of functions g1, g2,..., gk in I such that their sum is a function in I that never becomes zero. Multiplying this function by its reciprocal will imply that the identically 1 function is in I. Which implies I=C[0,1]

Hence the J(y)'s are the only maximal ideals in C[0,1] (?)

Thank you for reading the whole thing.

Your solution is almost correct, but there are a few minor errors. Let me clarify and correct them for you.

First of all, your definition of J(y) is not quite accurate. It should be J(y) = {f in C[0,1] / f(y) = 0} where y is fixed in [0,1]. This means that J(y) is the set of all continuous functions on [0,1] that vanish at the point y. This is indeed an ideal of C[0,1], which you have correctly shown using the two properties.

Next, you have correctly shown that C[0,1]/J(y) is a field, but you have not shown that it is isomorphic to the field of real numbers. This is an important step, as it shows that J(y) is a maximal ideal. You can do this by defining a map from C[0,1]/J(y) to the real numbers, where the image of f in C[0,1]/J(y) is f(y). This map is well-defined and is an isomorphism, which proves that C[0,1]/J(y) is isomorphic to the field of real numbers.

To show that there are no other maximal ideals, your idea is correct, but the execution is not entirely accurate. Here is a simpler way to prove this:

Assume that there exists an ideal I of C[0,1] such that I is not {0} and I is not a subset of any J(y). This means that there exists a function g in I such that g is not in any J(y). This implies that g is non-zero at every point in [0,1].

Now consider the set J(g) = {f in C[0,1] / fg = 0}. This is an ideal of C[0,1], since if f and h are in J(g), then fgh = 0 and f+h is in J(g). Also, g is not in J(g) since g is non-zero at every point in [0,1]. This means that J(g) is a proper ideal of C[0,1] that strictly contains I.

However, we have already shown that the J(y)'s are the only maximal ideals in C[0,1]. This means that J(g) cannot be a maximal ideal, which contradicts our assumption that I is

## What is a maximal ideal?

A maximal ideal is a proper subset of a ring that is closed under addition and multiplication, and is maximal with respect to this property. In other words, it is an ideal that cannot be properly contained in another ideal.

## How do you classify all maximal ideals on a ring?

There is no one definitive method for classifying all maximal ideals on a ring. However, a common approach is to first identify all the prime ideals on the ring, and then determine which of these are also maximal ideals.

## What are some properties of maximal ideals?

Maximal ideals have several important properties, including being prime, being relatively prime, and being radical. They are also useful for constructing quotient rings.

## Can a ring have more than one maximal ideal?

Yes, a ring can have multiple maximal ideals. In fact, a ring can have infinitely many maximal ideals if it is not a field.

## How do maximal ideals relate to other types of ideals?

Maximal ideals are a special type of ideal, and are closely related to other types of ideals such as prime ideals, principal ideals, and radical ideals. In some cases, maximal ideals can be used to construct these other types of ideals.

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